Suppose that you increase the instantaneous speed of a moving object, but you do not change its direction at that moment, nor do you change its acceleration. What effect does this have on the curvature of its path at that instant?
The answer is "It decreases", I thought it remains same as curvature doesn't depend on speed. Could someone please explain?
Suppose that your curve $\mathcal{c}$ is parametrized by $\gamma(t)$. The curvature as a function of the parameter $t$ is given by $$ \kappa(t) = \frac{\lvert\lvert \gamma^{\prime}(t) \times \gamma^{\prime\prime}(t) \rvert \rvert}{\lvert \lvert \gamma^{\prime}(t) \rvert\rvert^{3}}.$$
If you somehow managed to increase the speed, i.e. scaled the tangent vector $\gamma^{\prime}(t)$ by a constant $c > 1$ without changing the acceleration (so that somehow $\gamma^{\prime\prime}(t)$ didn't scale by the same factor), then the curvature $\kappa(t)$ would have to decrease as the denominator would have more factors of the constant $c > 1$ than the numerator would.
Note that if you (far more plausibly) uniformly scaled a curve $\gamma(t)$ by a constant factor $c$, then both $\gamma^{\prime}(t)$ and $\gamma^{\prime\prime}(t)$ would scale by the same factor $c$. If $c < 1$, then the curvature at every point would increase, while if $c >1$, the curvature $\kappa(t)$ would decrease. Note that this is relatively easy to see from the formula of $\kappa$ as the denominator of $\kappa$ would collect more factors of the constant $c$ than the numerator of $\kappa$ does. (In this instance, think of what happens to a circle of radius $r$ if one scales the radius by a factor $c > 1$: The curvature of a circle with radius $R > r$ is smaller than the curvature of the circle with radius $r$. Scaling everything up by a constant factor will increase the arc length required for the unit tangent to turn an indicated amount.)
For what it is worth, this--changing the speed without changing the acceleration--is a very interesting and potentially problematic way to think about curvature. The curvature of a curve $\mathcal{C}$ at a regular point $P$ is independent of how you decide to parametrize $\mathcal{C}$, but it doesn't make any real sense to change the speed without also having the acceleration change.